Problem:
Prove the proposition:
There exist larger operators that cannot be constructed
by tensoring smaller ones.
Solution:
We'll show by example. Consider the example unitary matrix
$$
C_{NOT} \eql
\mat{1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0}
$$
This happens to be one of the most important unitary matrices,
and we'll have more to say about this later. But for now let's
see what happens when we try to construct \(C_{NOT}\)
using two single-qubit unitaries:
$$
\mat{a_1 & b_1\\ c_1 & d_1}
\otimes
\mat{a_2 & b_2\\ c_2 & d_2}
\eql
\mat{1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0}
$$
That is,
$$
\mat{a_1a_2 & a_1b_2 & b_1a_2 & b_1b_2\\
a_1c_2 & a_1d_2 & b_1c_2 & c_1d_2\\
c_1a_2 & c_1b_2 & d_1a_2 & d_1b_2\\
c_1c_2 & c_1d_2 & d_1c_2 & d_1d_2}
\eql
\mat{1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0}
$$
The very first entry shows that \(a_1a_2=1\) and so neither
can be 0. Which means, for the next entry \(a_1b_2\) to be 0,
\(b_2=0\). This will make \(d_1b_2=0\).
Aside: we assumed that the product of two non-zero complex numbers can't be 0. Do you see why this is true?
Problem:
Use outer-product forms to show that \(X = HZH\).
Solution:
$$\eqb{
HZH & \eql &
\isqt{1} \parenl{ \otr{0}{0} + \otr{1}{0} + \otr{0}{1} - \otr{1}{1} }
\: \parenl{ \otr{0}{0} - \otr{1}{1} } \:H\\
& \eql &
\isqt{1} \parenl{ \otr{0}{0} + \otr{1}{0} - \otr{0}{1} + \otr{1}{1} } \: H \\
& \eql &
\frac{1}{2} \parenl{ \otr{0}{0} + \otr{1}{0} - \otr{0}{1} + \otr{1}{1} }
\: \parenl{ \otr{0}{0} + \otr{1}{0} + \otr{0}{1} - \otr{1}{1} } \\
& \eql &
\frac{1}{2} \parenl{ 2 \otr{1}{0} + 2 \otr{0}{1} } \\
& \eql &
X
}$$
The key is to look for cross-product terms where you can use
orthogonality to either zero-out or simplify (when inner-product = 1),
as in:
Problem:
Show without converting to matrices that,
for \(\cnot\), the tensored 1-qubit outer-product
form
$$
\cnot \eql
\otr{0}{0} \otimes I + \otr{1}{1} \otimes X
$$
is equivalent to
$$
\cnot \eql
\otr{00}{00} + \otr{01}{01} + \otr{10}{11} + \otr{11}{10}
$$
Solution:
$$\eqb{
\cnot
& \eql &
\otr{0}{0} \otimes I + \otr{1}{1} \otimes X \\
& \eql &
\otr{0}{0} \otimes \parenl{ \otr{0}{0} + \otr{1}{1} }
+ \otr{1}{1} \otimes \parenl{ \otr{0}{1} + \otr{1}{0} }\\
& \eql &
\otr{0}{0} \otimes \otr{0}{0} \plss \otr{0}{0} \otimes \otr{1}{1}
\plss \otr{1}{1} \otimes \otr{0}{1} \plss \otr{1}{1} \otimes \otr{1}{0}\\
& \eql &
\kt{0}\kt{0} \br{0}\br{0} \plss \kt{0}\kt{0} \br{1}\br{1}
\plss \kt{1}\kt{1} \br{0}\br{1} \plss \kt{1}\kt{1} \br{1}\br{0}\\
& \eql &
\otr{00}{00} + \otr{01}{01} + \otr{10}{11} + \otr{11}{10}
}$$
Problem:
Evaluate whether the 2-qubit state
$$
\ksi \eql \smf{\sqrt{3}}{2\sqrt{2}} \kt{00}
- \smf{\sqrt{3}}{2\sqrt{2}} \kt{01}
+ \smf{1}{2\sqrt{2}} \kt{10}
- \smf{1}{2\sqrt{2}} \kt{11}
$$
is entangled.
Solution:
Here, we see that
$$
a_{00} a_{11} - a_{01} a_{10}
\eql
\smf{\sqrt{3}}{2\sqrt{2}}
\parenl{ - \smf{1}{2\sqrt{2}} }
\; - \;
\parenl{- \smf{\sqrt{3}}{2\sqrt{2}} }
\smf{1}{2\sqrt{2}}
\eql 0
$$
Which makes this separable. Some factoring effort shows that
$$
\ksi
\eql
\parenl{ \smf{\sqrt{3}}{2} \kt{0} + \smf{1}{2} \kt{1} }
\otimes
\parenl{ \isqts{1} \kt{0} - \isqts{1} \kt{1} }
$$